Question

The side $ \mathrm{AB}$ of a parallelogram $ \mathrm{ABCD}$is produced to any point $ \mathrm{P}. $A line through $ \mathrm{A} $ and parallel to $ \mathrm{CP}$ meets $ \mathrm{CB}$ produced at $ \mathrm{Q} $ and then parallelogram $ \mathrm{PBQR}$ is completed (see Fig.) . Show that $ \mathrm{ar}\left(\mathrm{ABCD}\right) = \mathrm{ar}\left(\mathrm{PBQR}\right)$ . [Hint : Join $ \mathrm{AC} \mathrm{and} \mathrm{QP}$. Now compare $ \mathrm{ar}\left(\mathrm{ACQ}\right) \mathrm{and} \mathrm{ar}\left(\mathrm{APQ}\right) $ .]

D LN\, 

Answer 1

Given:

1. The side $\mathrm{AB}$ of a parallelogram $\mathrm{ABCD}$ is produced to any point $\mathrm{P}.$
2. Line through $\mathrm{A}$and parallel to $\mathrm{CP}$ meets $\mathrm{CB}$ produced at $\mathrm{Q}$and then the parallelogram $\mathrm{PBQR}$ is completed
3. Join $\mathrm{AC}$ and $\mathrm{PQ}$. Now compare $\mathrm{ar}\left(\mathrm{ACQ}\right)\mathrm{and}\mathrm{ar}\left(\mathrm{APQ}\right)$

To find: Prove that : $\mathrm{ar}\left(\mathrm{ABCD}\right)=\mathrm{ar}\left(\mathrm{PBQR}\right)$

Step 1: Prove that : $\mathrm{AC}\mathrm{and}\mathrm{QP}$ are diagonals $\mathrm{ABCD}\mathrm{and}\mathrm{PBQR}\mathit{.}$

$\mathrm{AC}\mathrm{and}\mathrm{PQ}$are joined.

$\mathrm{Ar}\left(\mathrm{\Delta ACQ}\right)=\mathrm{ar}\left(\mathrm{\Delta APQ}\right)$(On the same base $\mathrm{AQ}$ and between the same parallel lines $\mathrm{AQ}\mathrm{and}\mathrm{CP}$)

$=>\mathrm{ar}\left(\mathrm{\Delta ACQ}\right)-\mathrm{ar}\left(\mathrm{\Delta ABQ}\right)=\mathrm{ar}\left(\mathrm{\Delta APQ}\right)-\mathrm{ar}\left(\mathrm{\Delta ABQ}\right)$

$=>\mathrm{ar}(△\mathrm{ABC})=\mathrm{ar}(△\mathrm{QBP})—————\left(1\right)$

$\mathrm{AC}\mathrm{and}\mathrm{QP}$ are diagonals $\mathrm{ABCD}\mathrm{and}\mathrm{PBQR}$.

Step 2: Prove that $\mathrm{ar}\left(\mathrm{ABCD}\right)=\mathrm{ar}\left(\mathrm{PBQR}\right)$

Hence,

$\mathrm{ar}\left(\mathrm{ABC}\right)=\frac{1}{2}\times \mathrm{ar}\left(\mathrm{ABCD}\right)—————\left(2\right)$

$\mathrm{ar}\left(\mathrm{QBP}\right)=\frac{1}{2}\times \mathrm{ar}\left(\mathrm{PBQR}\right)—————\left(3\right)$

From above equation (2) and (3), we get

$$\begin{gathered} \frac{1}{2}\mathrm{ar}\left(\mathrm{ABCD}\right)=\frac{1}{2}\mathrm{ar}\left(\mathrm{PBQR}\right) \\ =>\mathrm{ar}\left(\mathrm{ABCD}\right)=\mathrm{ar}\left(\mathrm{PBQR}\right) \end{gathered}$$

Hence $\mathbf{ar}\mathbf{\left(}\mathbf{ABCD}\mathbf{\right)}\mathbf{}\mathbf{=}\mathbf{}\mathbf{ar}\mathbf{\left(}\mathbf{PBQR}\mathbf{\right)}$ is proved